Communication Systems • Topic 1 of 3

Elements of a Communication System

Communication is the act of transmitting information from one place to another. Every communication system, whether a simple telephone call or a satellite TV link, can be reduced to three essential blocks: a transmitter, a communication channel and a receiver. The information originates at a source and is delivered to a user (destination).

  • Transmitter: processes the message signal so it becomes suitable for transmission over the channel and for radiation by the antenna. It typically contains a transducer, a modulator and an amplifier.
  • Channel: the physical medium that carries the signal from transmitter to receiver — a pair of wires, a coaxial cable, an optical fibre, or free space (for radio waves).
  • Receiver: extracts the original message signal from the received (often weak and distorted) signal through amplification, demodulation and detection.

Transducer: a device that converts one form of energy into another. A microphone converts sound (pressure variations) into an electrical signal; a loudspeaker does the reverse. Communication electronics works only with electrical signals, so transducers sit at the input of the transmitter and the output of the receiver.

Analog vs digital signals. An analog signal is continuous in both time and amplitude — speech and music waveforms are analog. A digital signal takes only discrete values (usually two levels, $0$ and $1$). Digital communication is more robust against noise, can be regenerated cleanly by repeaters, and is easily encrypted, which is why modern systems are overwhelmingly digital.

Signal degradation in the channel. Three effects act on a signal as it travels:

  • Attenuation: the loss of signal strength (power) as it propagates, caused by absorption and spreading in the medium. It is measured in decibels (dB).
  • Amplification: increasing the amplitude (and power) of a signal using an electronic amplifier. Repeaters placed along the channel pick up, amplify and retransmit the signal to compensate for attenuation.
  • Noise: unwanted, random electrical signals that get added to the message during transmission or processing. Noise sets the ultimate limit on how weak a usable signal can be.

Gain in decibels. Because power levels span huge ranges, gain and loss are expressed logarithmically: $\text{Gain (dB)} = 10\log_{10}\dfrac{P_o}{P_i}$, where $P_o$ and $P_i$ are output and input powers. A gain of $+20\,\text{dB}$ means the power increased $100$ times; an attenuation of $-30\,\text{dB}$ means the power fell to one-thousandth.

Bandwidth of a signal. Every message occupies a band of frequencies. The bandwidth is the range $\Delta f = f_{\max} - f_{\min}$ that must be transmitted faithfully. Typical message bandwidths are: speech for telephony about $2800\,\text{Hz}$ ($300$ Hz to $3100$ Hz); high-quality music up to $20\,\text{kHz}$ (audible range $20\,\text{Hz}$ to $20\,\text{kHz}$); a video signal about $4.2\,\text{MHz}$; and a composite TV signal (video + audio) about $6\,\text{MHz}$.

Bandwidth of transmission media. The carrying capacity of a medium increases with its usable frequency range. Twisted-pair telephone wire handles a few hundred kHz; coaxial cable supports up to about $750\,\text{MHz}$; and an optical fibre, working at optical frequencies near $10^{14}\,\text{Hz}$, offers an enormous bandwidth (tens of GHz to THz), which is why fibre carries the bulk of the world's high-speed data with very low attenuation.

Block diagram: source to transmitter to channel to receiver to destinationInformationSourceTransmitterChannelReceiverUser /DestinationNoise enters hereElements of a Communication System
1
Worked Example
An amplifier raises the signal power from $1\,\mu\text{W}$ to $100\,\text{mW}$. Express its gain in decibels.
Solution
  1. Input power $P_i = 1\,\mu\text{W} = 10^{-6}\,\text{W}$; output power $P_o = 100\,\text{mW} = 0.1\,\text{W}$.
  2. Power ratio $\dfrac{P_o}{P_i} = \dfrac{0.1}{10^{-6}} = 10^{5}$.
  3. Gain $= 10\log_{10}(10^{5}) = 10 \times 5 = 50\,\text{dB}$.

Answer: The amplifier has a gain of $50\,\text{dB}$.

2
Worked Example
A signal of $2\,\text{mW}$ is fed into a cable. Over its length the cable attenuates the signal by $20\,\text{dB}$. Find the output power.
Solution
  1. Attenuation in dB: $-20 = 10\log_{10}\dfrac{P_o}{P_i}$.
  2. So $\log_{10}\dfrac{P_o}{P_i} = -2 \Rightarrow \dfrac{P_o}{P_i} = 10^{-2} = 0.01$.
  3. $P_o = 0.01 \times 2\,\text{mW} = 0.02\,\text{mW} = 20\,\mu\text{W}$.

Answer: The output power is $20\,\mu\text{W}$ (the signal has dropped to one-hundredth of its input).

3
Worked Example
The audible range of frequencies for the human ear is $20\,\text{Hz}$ to $20\,\text{kHz}$. What bandwidth is needed to transmit a high-fidelity music signal covering this range?
Solution
  1. Bandwidth $\Delta f = f_{\max} - f_{\min}$.
  2. $f_{\max} = 20\,\text{kHz} = 20000\,\text{Hz}$, $f_{\min} = 20\,\text{Hz}$.
  3. $\Delta f = 20000 - 20 = 19980\,\text{Hz} \approx 20\,\text{kHz}$.

Answer: A bandwidth of about $20\,\text{kHz}$ is required for high-fidelity music.

4
Worked Example
Telephone-quality speech is transmitted over the band $300\,\text{Hz}$ to $3100\,\text{Hz}$. State the bandwidth and explain why this is narrower than for music.
Solution
  1. Bandwidth $\Delta f = 3100 - 300 = 2800\,\text{Hz}$.
  2. Intelligible speech is concentrated in the mid-frequency band; frequencies below $300\,\text{Hz}$ and above $3100\,\text{Hz}$ add little to recognising words.
  3. Restricting the band to $2.8\,\text{kHz}$ keeps speech understandable while saving channel capacity, so many calls share a medium.

Answer: The bandwidth is $2800\,\text{Hz}$; speech needs less band than music because intelligibility depends mainly on mid-range frequencies.

5
Worked Example
A composite TV signal occupies a bandwidth of about $6\,\text{MHz}$. If a coaxial cable can support frequencies up to $750\,\text{MHz}$, roughly how many such TV channels can it carry simultaneously?
Solution
  1. Usable cable bandwidth $\approx 750\,\text{MHz}$ (treating the full band as available).
  2. Each channel needs $6\,\text{MHz}$.
  3. Number of channels $\approx \dfrac{750}{6} = 125$.

Answer: About $125$ TV channels can be carried at once on the coaxial cable.

6
Worked Example
Two amplifiers with gains $30\,\text{dB}$ and $20\,\text{dB}$ are cascaded, while a cable of $15\,\text{dB}$ loss connects them. Find the net gain in dB and the overall power ratio.
Solution
  1. In dB, gains add and losses subtract: net gain $= 30 + 20 - 15 = 35\,\text{dB}$.
  2. Net gain $= 10\log_{10}\dfrac{P_o}{P_i} = 35 \Rightarrow \log_{10}\dfrac{P_o}{P_i} = 3.5$.
  3. Power ratio $\dfrac{P_o}{P_i} = 10^{3.5} \approx 3162$.

Answer: Net gain $= 35\,\text{dB}$, an overall power amplification of about $3162$ times.

Key Points

  • A communication system has three core blocks — transmitter, channel and receiver — linking an information source to the user.
  • A transducer converts energy forms (e.g. microphone: sound to electrical; loudspeaker: electrical to sound).
  • Analog signals are continuous; digital signals are discrete and more noise-resistant; gain/loss in dB is $10\log_{10}(P_o/P_i)$.
  • Attenuation weakens a signal, amplification strengthens it (repeaters do both), and noise sets the limit on detection.
  • Bandwidth: speech $\approx 2.8\,\text{kHz}$, music $\approx 20\,\text{kHz}$, video $\approx 4.2\,\text{MHz}$, TV $\approx 6\,\text{MHz}$; media: wire (kHz), coaxial ($\sim 750\,\text{MHz}$), optical fibre (huge).
Tap an option to check your answer0 / 4
Q1.Which device converts a non-electrical message into an electrical signal at the transmitter input?
Explanation: A transducer converts one form of energy into another; a microphone (a transducer) turns sound into an electrical signal.
Q2.The approximate bandwidth required to transmit a standard composite TV signal is:
Explanation: A composite (video + audio) TV signal occupies a bandwidth of about $6\,\text{MHz}$.
Q3.An amplifier increases signal power by a factor of $1000$. Its gain in decibels is:
Explanation: Gain $= 10\log_{10}(1000) = 10 \times 3 = 30\,\text{dB}$.
Q4.Which transmission medium offers the largest bandwidth and lowest attenuation for long-distance high-speed data?
Explanation: Optical fibre operates near $10^{14}\,\text{Hz}$, giving an enormous bandwidth with very low loss, ideal for high-speed long-haul links.